# Simple Harmonic Motion, platform and box

## Homework Statement

A platform moves up and down in SHM, with amplitude 0.035 m. Resting on top of the platform is a block of wood. What is the shortest period of motion for the platform so that the block will remain in constant contact with it?

## Homework Equations

a(t)=-A$$\omega$$^2cos($$\omega$$t+phase constant)
amax=A$$\omega$$

## The Attempt at a Solution

I didn't see how I could possibly use the first one with so many unknowns so I used amax=A$$\omega$$ and set amax=9.8 figuring that the amount couldn't be more than gravity otherwise the block and platform would separate (maybe I'm wrong in this).
And so I set 9.8=.035$$\omega$$ and solved (getting 280). The I used $$\omega$$=2$$\pi$$f and solving for f (getting 140/$$\pi$$) and then using f=1/T and solved for T getting .0224 which was incorrect.

## Answers and Replies

ehild
Homework Helper
Check your formula for amax, it is not correct.

ehild

Oh, it came out funny. It's supposed to be amax=A x omega. Is this still incorrect? If it is, I really need to throw a fit about a refund.

ehild
Homework Helper
No, see the previous line. a must be m/s2, yours is m/s.

ehild